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3 years ago

6

A population of frogs in a pond increases at an annual rate of 22%. If there are 100 frogs in the pond, which equation gives you

a model for how many frogs there will be in time t years?

2 answers:

GarryVolchara [31]3 years ago

7 0

Answer:

y = 100 (1.22)^t

Step-by-step explanation:

this is exponential growth

y = a (1+r) ^t

where a is the initial amount

r is the percent growth

t is the time in years

y= 100(1 + .22)^t

y = 100 (1.22)^t

Papessa [141]3 years ago

4 0

Answer:

Step-by-step explanation:

Y=100(1.22)^t

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Tonya and Leo each bought a cell phone at the same time. The trade-in values, in dollars, of the cell phones are modeled by the

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Answer:

The answer is Tonya's phone had the greater initial trade-in value.

Leo's phone decreases at an average rate slower than the trade in value of Tonya's phone.

Step-by-step explanation:

Given

Tonya

$f(x) = 490 * 0.88^x$

Leo

$x \to g(x)$

$0 \to 480$

$2 \to 360$

$4 \to 470$

Solving (a): The phone with greater initial value

The initial value is when x = 0. So, we have:

$f(x) = 490 * 0.88^x$

$f(0) = 490 * 0.88^0$

$f(0) = 490 * 1$

$f(0) = 490$

From Leo's table

$g(0) = 480$

By comparison;

$f(0) > g(0)$

i.e.

$490 > 480$

<em>So: Tonya's had the greater initial trade-in value</em>

Solving (b): The phone with lesser rate

An exponential function is:

$y = ab^x$

Where:

$b \to rate$

For Tonya

$b = 0.88$

For Leo, we have:

$(x_1,y_1) = (0,480)$

$(x_2,y_2) = (2,360)$

So, the equation becomes:

$y = ab^x$

$480 = ab^0$ and $360 = ab^2$

Solving $480 = ab^0$ , we have:

$480 = a * 1$

$480 = a$

$a= 480$

$360 = ab^2$ becomes

$360 = 480 * b^2$

Divide both sides by 480

$0.75 = b^2$

Take square roots

$0.87 = b$

$b=0.87$ -- Leo's rate

By comparison; Leo's rate is slower i.e. 0.87 < 0.88

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